NMTC Inter Level Problem 12

Since $15^2+20^2=25^2$ , then we have a right triangle with legs of lengths 15 and 20, and hypotenuse of length 25.

From this, it follows that if set the 15 as the base, the altitude of the triangle is 20, and if we set the 20 as the base, the altitude of the triangle is 15. We also take note that the area of the right triangle is $A=\dfrac{1}{2}bh=\dfrac{1}{2}(15)(20)=150.$

What if we set the hypotenuse 25 as the base? We can find the length of the corresponding altitude by the area formula.

$\dfrac{1}{2}(\text{length of the hypotenuse})(\text{altitude to the hypotenuse})=150$ $\dfrac{1}{2}(25)(h_{\text{hyp}})=150$ Therefore, $h_{\text{hyp}}=12.$ But among the lengths of altitudes that we have (12, 15, 20), 12 is the smallest. Therefore, the answer is $\boxed{12}$ .